Difference between revisions of "2004 AIME II Problems/Problem 8"

Latest revision as of 17:36, 28 June 2025

Problem

How many positive integer divisors of $2004^{2004}$ are divisible by exactly 2004 positive integers?

Solution 1

The prime factorization of 2004 is $2^2\cdot 3\cdot 167$ . Thus the prime factorization of $2004^{2004}$ is $2^{4008}\cdot 3^{2004}\cdot 167^{2004}$ .

We can count the number of divisors of a number by multiplying together one more than each of the exponents of the prime factors in its prime factorization. For example, the number of divisors of $2004=2^2\cdot 3^1\cdot 167^1$ is $(2+1)(1+1)(1+1)=12$ .

A positive integer divisor of $2004^{2004}$ will be of the form $2^a\cdot 3^b\cdot 167^c$ . Thus we need to find how many $(a,b,c)$ satisfy

$(a+1)(b+1)(c+1)=2^2\cdot 3\cdot 167.$

We can think of this as partitioning the exponents to $a+1,$ $b+1,$ and $c+1$ . So let's partition the 2's first. There are two 2's so this is equivalent to partitioning two items in three containers. We can do this in ${4 \choose 2} = 6$ ways. We can partition the 3 in three ways and likewise we can partition the 167 in three ways. So we have $6\cdot 3\cdot 3 = \boxed{54}$ as our answer.

Solution 2 (bash)

Clearly we need to find a group of numbers that multiply to 2004. We can list them all out since we know that 2004 is only $167 \cdot 2^2 \cdot 3$ .

167, 2, 2, 3

4, 3, 167

12, 167

4, 501

2, 1002

2, 3, 334

2, 2, 501*

6, 2, 167

3, 668

6, 334

2004*

To begin, the first multiple doesn't work because there are only 3 prime divisors of 2004. We can apply all multiples because the prime factorization of $2004^{2004}$ is $2^{4008} \cdot 3^{2004} \cdot 167^{2004}.$ Every multiple has six ways of distributing numbers to become powers of 167, 3, and 2, except for the ones with a star. For a single power of 2004, we have three choices (2, 3, and 167) to give a power of 2003 to. For 2, 2, 501, there are three choices to give a power of 500 to and the rest get a power of 1.

Therefore we have $8 \cdot 6$ normal multiples and $3 \cdot 2$ "half" multiples. Sum to get $\boxed{54}$ as our answer.

Solution 3

we want $167^a \cdot 3^b \cdot 2^c$ such that this number has exactly 2004 positive factors.

Case 1: $(a + 1)(b + 1)(c + 1) = 2004$ , and no variable equals 0.

$a + 1, b + 1, c + 1$ can either be {167, 2, 6}, {501, 2, 2}, {334, 3, 2}, or {167, 4, 3}. They can be permuted in $6 + 6 + 3 + 6 =$ 15 ways.

Case 2: $(x + 1)(y + 1) = 2004$ , where $x$ and $y$ are two of the variables out of the three ${a, b, c}$ and none of them equal 0.

There are $12$ ways to choose a value of $x$ and 1 way to choose a value of $y$ . There are 3 ways to choose the variables $x$ and $y$ out of $a, b, c$ , giving $36$ ways.

Final case: $(x + 1) = 2004$

$x$ must be $2003$ . This can happen in $3$ ways .

The answer is $15 + 36 + 3 = \boxed{54}.$

~grogg007

@@ Line 1: / Line 1: @@
 == Problem ==
-How many [[positive integer]] [[divisor]]s of <math> 2004^{2004} </math> are [[divisibility | divisible]] by exactly 2004 positive integers?
+How many positive integer divisors of <math>2004^{2004}</math> are divisible by exactly 2004 positive integers?
-== Solution ==
+== Solution 1 ==
 The [[prime factorization]] of 2004 is <math>2^2\cdot 3\cdot 167</math>.  Thus the prime factorization of <math>2004^{2004}</math> is <math>2^{4008}\cdot 3^{2004}\cdot 167^{2004}</math>.
@@ Line 11: / Line 11: @@
 <center><math>(a+1)(b+1)(c+1)=2^2\cdot 3\cdot 167.</math></center>
-We can think of this as [[partition]]ing the exponents to <math>a+1,</math>  <math>b+1,</math> and  <math>c+1</math>.  So let's partition the 2's first.  There are two 2's so this is equivalent to partitioning two items in three containers.  We can do this in <math>{4 \choose 2} = 6</math> ways.  We can partition the 3 in three ways and likewise we can partition the 167 in one way.  So we have <math>6\cdot 3\cdot 3 = 054</math> as our answer.
+We can think of this as [[partition]]ing the exponents to <math>a+1,</math>  <math>b+1,</math> and  <math>c+1</math>.  So let's partition the 2's first.  There are two 2's so this is equivalent to partitioning two items in three containers.  We can do this in <math>{4 \choose 2} = 6</math> ways.  We can partition the 3 in three ways and likewise we can partition the 167 in three ways.  So we have <math>6\cdot 3\cdot 3 = \boxed{54}</math> as our answer.
+==Solution 2 (bash)==
+Clearly we need to find a group of numbers that multiply to 2004. We can list them all out since we know that 2004 is only <math>167 \cdot 2^2 \cdot 3</math>.
+, 2, 2, 3
+, 3, 167
+, 167
+, 501
+, 1002
+, 3, 334
+, 2, 501*
+, 2, 167
+, 668
+, 334
+*
+To begin, the first multiple doesn't work because there are only 3 prime divisors of 2004. We can apply all multiples because the prime factorization of <math>2004^{2004}</math> is <math>2^{4008} \cdot 3^{2004} \cdot 167^{2004}.</math> Every multiple has six ways of distributing numbers to become powers of 167, 3, and 2, except for the ones with a star.
+For a single power of 2004, we have three choices (2, 3, and 167) to give a power of 2003 to.
+For 2, 2, 501, there are three choices to give a power of 500 to and the rest get a power of 1.
+Therefore we have <math>8 \cdot 6</math> normal multiples and <math>3 \cdot 2</math> "half" multiples. Sum to get <math>\boxed{54}</math> as our answer.
+==Solution 3==
+we want <math>167^a \cdot 3^b \cdot 2^c</math> such that this number has exactly 2004 positive factors.
+'''Case 1:''' <math>(a + 1)(b + 1)(c + 1) = 2004</math>, and no variable equals 0.
+<math>a + 1, b + 1, c + 1</math> can either be {167, 2, 6}, {501, 2, 2}, {334, 3, 2}, or {167, 4, 3}. They can be permuted in <math>6 + 6 + 3 + 6 = </math>15 ways.
+'''Case 2: <math>(x + 1)(y + 1) = 2004</math>''', where <math>x</math> and <math>y</math> are two of the variables out of the three <math>{a, b, c}</math> and none of them equal 0.
+There are <math>12</math> ways to choose a value of <math>x</math> and 1 way to choose a value of <math>y</math>. There are 3 ways to choose the variables <math>x</math> and <math>y</math> out of  <math>a, b, c</math>, giving <math>36</math> ways.
+'''Final case:''' <math>(x + 1) = 2004</math>
+<math>x</math> must be <math>2003</math>. This can happen in <math>3</math> ways .
+The answer is <math>15 + 36 + 3 = \boxed{54}.</math>
+~[[User:grogg007|grogg007]]
 == See also ==
@@ Line 18: / Line 84: @@
 [[Category:Intermediate Number Theory Problems]]
 [[Category:Intermediate Combinatorics Problems]]
+{{MAA Notice}}

2004 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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